Physics 137 - Introduction to the Atmosphere and Weather                                                                           Homework Key - Assignment #7
Revised Fall 1999

7-QT2.   Are the bases of convective clouds generally higher during the day or the night?  Explain.

   At night, because of lower temperatures at the surface, the difference between the dew point and the surface temperature is smaller than during the day. Therefore an air parcel need not rise so high for it to become saturated. If convection is able to occur, the cloud bases will therefore be lower at night.

7-QT4.   Use Fig. 5.11, p.111 (Chapter 4), to help you explain why the bases of cumulus clouds, which form from rising thermals during the summer, increase in height above the surface as you move due west of a line that runs north-south through central Kansas.

   West of the described line, average summer water vapor pressure decreases rapidly to the west and therefore the dew point and relative humidity drops to the west. Since the height of convective cloud bases is proportional to the difference in the temperature and the dew point (see the box on p.172), the height of cumulus bases increases to the west.

7-QT5.   For least polluted conditions, what would be the best time of day for a farmer to burn agricultural debris?

    For least pollution the farmer should burn his debris at that time of day most favorable for convection, i.e., at that time when the atmosphere is least stable. That tends to be the case at the warmest time of day. Thus the farmer should burn his debris at the warmest time of day, probably from mid to late afternoon.

7-PE2.   If the height of a cumulus cloud is 1000 meters above the surface, and the dew point at the earth's surface beneath the cloud is 20EC, determine the air temperature at the earth's surface beneath the cloud.

    The dew-point lapse rate is 2EC/km, hence the temperature at the cloud base is 18EC. Since the dry adiabatic rate is 10EC/km, the temperature at the surface is therefore 18EC+10EC=28EC.

7-PE3.  The condensation level over New Orleans, Louisiana, on a warm muggy afternoon is 2000 feet. If the dew point of the rising air at this level is 73EF, what is the approximate dew point and air temperature at the surface?  Determine the surface relative humidity. (Hint: See Chapter 5, p.117.)

    Since the dew-point lapse rate is 2EC/km=1EF/1000 ft, the dew point at the surface is T_dew=73EF+2EF=75EF. Since the dry adiabatic rate is approximately 5.5EF/1000 ft, the temperature at the surface is T=73EF+2×5.5EF=84EF. Hence the vapor pressure at the surface, which is equal to the saturation vapor pressure at 75EF is (from Table 1 on p.117) e=29.6 mb. The saturation vapor pressure at the surface (84EF) is (from Table 1 on p.117) e_s=39.8 mb. Thus the relative humidity is RH=e/e_s=29.6 mb/39.8 mb=74%.

7-PE5.  In Fig.7.26, above, a radiosonde is released and sends back temperature data as shown in the diagram. (This is the environment temperature.) (a) Calculate the environmental lapse rate from the surface up to 3000 m. (b) What type of atmospheric stability does the sounding indicate?
        Suppose the wind is blowing from the west and a parcel of surface air with a temperature of 10°C and a dew point of 2°C begins to rise upward along the western (windward) side of the mountain. (c) What is the relative humidity of the air parcel at 0 m (pressure 1013 mb) before rising? (Hint:   See Chapter 5, p.126.) (d) As the air parcel rises, at approximately what elevation would condensation begin and a cloud start to form? (e) What is the air temperature and dew point of the rising air at the base of the cloud? (f) What is the air temperature and dew point of the rising air inside the cloud at an elevation of 3000 m? (Use moist adiabatic rate of 6°C/km.) (g) At an altitude of 3000 m, how does the air temperature inside the cloud compare with the air temperature outside the cloud, as measured by the radiosonde? What type of atmospheric stability (stable or unstable) does this suggest? Explain. (h) At an elevation of 3000 m, would you expect the cloud to continue to develop vertically? Explain. (i) What would be the name of the cloud that is forming?
        Suppose that a parcel of air inside the cloud descends from the top of the mountain at 3000 m (pressure 700 mb) down the eastern (leeward) side of the mountain to an elevation of 0 meters (pressure 1013 mb). (j) If the descending air warms at the dry adiabatic rate from the top of the mountain all the way down to 0m, what is the sinking air's temperature and dew point when it reaches 0m? (k) What would be the relative humidity of the sinking air at 0m? (Hint: See Chapter 5, p.117.) (l) What accounts for the sinking air being warmer at the base of the mountain on the eastern side? (m) Explain why the sinking air is drier (its dew point is lower) on the eastern side at 0 meters.

    (a) ELR = DT/h = [10EC-(-14EC)]/3000 m = 8EC/km. (b) 8EC is between the dry adiabatic rate and the moist adiabatic rate. Hence the atmosphere is conditionally stable. (c) The vapor pressure at the surface, which is equal to the saturation vapor pressure at the dew point of 2EC is (from Table 1 on p.117) e = 6.9 mb. The saturation vapor pressure at the surface (10EC) is (from Table 1 on p.117) e_s = 12.3mb. Thus the relative humidity is RH = e/e_s = 6.9 mb/12.3 mb = 56%. (d) Since the dry adiabatic rate and the dew-point lapse rate close at 8EC/km and since the temperature and the dew point differ by 8EC (10EC-2EC) at the surface, the height of the condensation level is approximately h = 1 km = 1000 m. (e) At the cloud base, the temperature will have dropped by 10EC from the surface, hence T_base= 0EC. (f) T(3000 m)=0EC-2000 m×6EC/1000 m = -12EC. (g) The inside temperature (-12EC) is higher than the outside temperature (-14EC). This suggests buoyancy for the rising moist parcel, i.e., conditional instability. (h) As explained in part (g) the parcel would continue to rise and therefore to develop vertically. (i) This would be a cumulus congestus cloud. (j) T_base= -12EC+3 km×10EC/km = 18EC, T_dew,base = -12EC+3 km×2EC/km = -6EC. (k) RH = e/e_s = 4.0 mb/21.0 mb = 19%. (l) Latent heat released during condensation causes the air to be warmer on the leeward side. (m) Water vapor removed from the air by precipitation on the windward side causes the air to be drier on the leeward side.